Calculate the ∆H for the following reaction: Sn + 2Cl2 🡪 SnCl4

Question

You are given these two equations:

Sn  +  Cl2 🡪 SnCl2  ∆H = -325 kJ  

SnCl2 + Cl2 🡪 SnCl4  ∆H =   -186 kJ  

Thermochemistry - Hess Law

Answer 1

Hi Priscilla,

In this problem, the strategy is to use Hess's law, which states that the change in enthalpy of a reaction can be determined by adding together known reaction enthalpies.

In the two equations given, we see that all of the species we need for the final equation are there: Sn, Cl₂, and SnCl₄. The only other species is SnCl₂, which is not in our target equation, so let's figure out how to cancel it out.

We can add equations by writing all of the reactants on the left side and products on the right, then cancelling out whatever is common on both sides. If we write:

(Sn + Cl₂) + (SnCl₂ + Cl₂) -> (SnCl₂) + (SnCl₄)

Then cancel common species out, we get our target equation:

Sn + 2Cl₂ -> SnCl₄

The Cl₂ species add up to 2Cl₂ since there are two on the left side.

Hess's law says that we can add together the enthalpies of these reactions to get the final enthalpy, so:

-325 kJ/mol + -186 kJ/mol = -511 kJ/mol

We did not need to flip any reactions or multiply by any factors in this problem, but it is something to look out for in other problems.

I hope this answer helped! Please feel free to respond with any more questions!

-Jacob P.

Answer 2

Sn + 2Cl₂ ==> SnCl₄ ... target equation

Using Hess' Law, we can use the other two equations/reactions that are given.

Sn + Cl₂ ==> SnCl₂ ∆H = -325 kJ

SnCl₂ + Cl₂ ==> SnCl₄ ∆H = -186 kJ

Adding them we get...

Sn + Cl₂ + SnCl₂ + Cl₂ ==> SnCl₂ + SnCl₄ ∆H = -325 + -186 (add the ∆H values)

Canceling common species, we end up with ..

Sn + 2Cl₂ ==> SnCl₄ the target equation.

∆Hrxn = -325 + -186 = -511 kJ